Treating Singularities

5.1.2 Treating Singularities

When the integrand or some of its low-order derivative is infinite at some point in or near the interval of integration, standard quadrature rules will not work well. It is not uncommon that a single step taken close to such a singular point will give a larger error than all other steps combined. In some cases a singularity can be completely missed by the quadrature rule.

If the singular points are known, then the integral should first be broken up into several pieces so that all the singularities are located at one (or both) ends of the interval [a, b].

Many integrals can then be treated by weighted quadrature rules, i.e., the singularity is incorporated into the weight function. Romberg’s method can be modified to treat integrals where the integrand has an algebraic endpoint singularity; see Sec. 5.2.2.

It is often profitable to investigate whether one can transform or modify the given problem analytically to make it more suitable for numerical integration. Some difficulties and possibilities in numerical integration are illustrated below in a series of simple examples.

Example 5.1.1.

In the integral

√ e x dx

the integrand is infinite at the origin. By the substitution x = t 2 we get

t I 2 =2 e dt,

which can be treated without difficulty. Another possibility is to use integration by parts:

3 3 0 3 3 0 The last integral has a mild singularity at the origin. If one wants high accuracy, then it is

advisable to integrate by parts a few more times before the numerical treatment.

Example 5.1.2.

Sometimes a simple comparison problem can be used. In

x −3 e x dx

the integrand is infinite near the left endpoint. If we write

x −3 e x −1−x− dx,

526Chapter 5. Numerical Integration the first integral can be computed analytically. The second integrand can be treated numeri-

cally. The integrand and its derivatives are of moderate size. Note, however, the cancellation in the evaluation of the integrand.

For integrals over an infinite interval one can try some substitution which maps the interval (0, ∞) to (0, 1), for example, t = e −x of t = 1/(1 + x). But in such cases one

must be careful not to introduce an unpleasant singularity into the integrand instead.

Example 5.1.3.

More general integrals of the form

2h

x −1/2 f (x) dx

need a special treatment, due to the integrable singularity at x = 0. Aformula which is exact for any second-degree polynomial f (x) can be found using the method of undetermined

coefficients. We set

1 2h √

x −1/2 f (x) dx ≈w 0 f( 0) + w 1 f (h) +w 2 f( 2h), 2h 0

and equate the left- and right-hand sides for f (x) = 1, x, x 2 . This gives

2 3 4 5 This linear system is easily solved, giving w 0 = 12/15, w 1 = 16/15, w 2 = 2/15.

Example 5.1.4.

Consider the integral

1+x 2 ) −4/3 dx.

If one wants five decimal digits in the result, then ∞

R is not negligible until R ≈ 10 . But one can expand the integrand in powers of x −1 and integrate termwise:

( 1+x 2 ) −4/3 dx =

x −8/3 ( 1+x −2 ) −4/3 dx

51 . −··· If this expansion is used, then one need only apply numerical integration to the interval

[0, 8]. With the substitution t = 1/(1 + x) the integral becomes

(t 2 + (1 − t) 2 ) −4/3 t 2/3 dt.

5.1. Interpolatory Quadrature Rules 527 The integrand now has an infinite derivative at the origin. This can be eliminated by making

the substitution t = u 3 to get

I (u 6 = 3 + (1 − u ) 2 ) −4/3 3u 4 du,

which can be computed with, for example, a Newton–Cotes’ method.